问题
I'm trying to write my own code for a 'friends-of-friends' algorithm. This algorithm acts on a set of 3d data points and returns the number of 'halos' in the dataset. Each halo is a collection of point whose distance is less than the linking length, b, the only parameter of the program.
Algorithmic Description: The FOF algorithm has a single free parameter called the linking length. Any two particles that are separated by a distance less than or equal to the linking length are called "friends". The FOF group is then defined by the set of particles for which each particle within the set is connected to every other particle in the set through a network of friends.
Set FOF group counter j=1.
For each particle, n, not yet associated with any group:
Assign n to group j, initialize a new member list, mlist, for group j with particle n as first entry,
Recursively, for each new particle p in mlist:
- Find neighbors of p lying within a distance less than or equal to the linking length, add to mlist those not already assigned to group j,
- Record mlist for group j, set j=j+1.
This is my attempt to code the algorithm. The only language I'm comfortable in doing this is Python. However, I need this code to be written in Fortran or make it faster. I really hope someone would help me.
First I generate a set of points that should mimic the presence of 3 halos:
import random
from random import *
import math
from math import *
import numpy
from numpy import *
import time
points = 1000
halos=[0,100.,150.]
x=[]
y=[]
z=[]
id=[]
for i in arange(0,points,1):
x.append(halos[0]+random())
y.append(halos[0]+random())
z.append(halos[0]+random())
id.append(i)
for i in arange(points,points*2,1):
x.append(halos[1]+random())
y.append(halos[1]+random())
z.append(halos[1]+random())
id.append(i)
for i in arange(points*2,points*3,1):
x.append(halos[2]+random())
y.append(halos[2]+random())
z.append(halos[2]+random())
id.append(i)
Then I coded the FOF algorithm:
x=array(x)
y=array(y)
z=array(z)
id=array(id)
t0 = time.time()
id_grp=[]
groups=zeros((len(x),1)).tolist()
particles=id
b=1 # linking length
while len(particles)>0:
index = particles[0]
# remove the particle from the particles list
particles.remove(index)
groups[index]=[index]
print "#N ", index
dx=x-x[index]
dy=y-y[index]
dz=z-z[index]
dr=sqrt(dx**2.+dy**2.+dz**2.)
id_to_look = where(dr<b)[0].tolist()
id_to_look.remove(index)
nlist = id_to_look
# remove all the neighbors from the particles list
for i in nlist:
if (i in particles):
particles.remove(i)
print "--> neighbors", nlist
groups[index]=groups[index]+nlist
new_nlist = nlist
while len(new_nlist)>0:
index_n = new_nlist[0]
new_nlist.remove(index_n)
print "----> neigh", index_n
dx=x-x[index_n]
dy=y-y[index_n]
dz=z-z[index_n]
dr=sqrt(dx**2.+dy**2.+dz**2.)
id_to_look = where(dr<b)[0].tolist()
id_to_look = list(set(id_to_look) & set(particles))
nlist = id_to_look
if (len(nlist)==0):
print "No new neighbors found"
else:
groups[index]=groups[index]+nlist
new_nlist=new_nlist+nlist
print "------> neigh-neigh", new_nlist
for k in nlist:
particles.remove(k)
At the end one ends up with a list of the halos in the list groups
This part of the code is a bit off topic but I thought it would be nice to show it to you. I am basically deleting all the groups with no particles, sorting them according to the number of particles and showing some properties.
def select(test,list):
selected = []
for item in list:
if test(item) == True:
selected.append(item)
return selected
groups=select(lambda x: sum(x)>0.,groups)
# sorting groups
groups.sort(lambda x,y: cmp(len(x),len(y)))
groups.reverse()
print time.time() - t0, "seconds"
mass=x
for i in arange(0,len(groups),1):
total_mass=sum([mass[j] for j in groups[i]])
x_cm = sum([mass[j]*x[j] for j in groups[i]])/total_mass
y_cm = sum([mass[j]*y[j] for j in groups[i]])/total_mass
z_cm = sum([mass[j]*z[j] for j in groups[i]])/total_mass
dummy_x_cm = [x[j]-x_cm for j in groups[i]]
dummy_y_cm = [y[j]-y_cm for j in groups[i]]
dummy_z_cm = [z[j]-z_cm for j in groups[i]]
dummy_x_cm = array(dummy_x_cm)
dummy_y_cm = array(dummy_y_cm)
dummy_z_cm = array(dummy_z_cm)
dr = max(sqrt(dummy_x_cm**2.+dummy_y_cm**2.+dummy_z_cm**2.))
dummy_x_cm = max(dummy_x_cm)
dummy_y_cm = max(dummy_y_cm)
dummy_z_cm = max(dummy_z_cm)
print i, len(groups[i]), x_cm, y_cm, z_cm,dummy_x_cm,dummy_y_cm,dummy_z_cm
回答1:
I think that you would be ill-advised to start off learning Fortran in the hope that the resulting code will be faster than your current implementation. It may, eventually, be, but I think you would be better advised to make your Python implementation as fast as you can before thinking of implementing in another language, especially a foreign language.
I write Fortran, and personally I think its performance pisses all over Python, but people who know about these things provide compelling arguments that Python+SciPy+Numpy can, if carefully crafted, match Fortran for speed in the computational kernels of many scientific/engineering programs. Don't forget that you haven't optimised your Python until all the cores on your computer are running red hot.
So:
1st - get a working implementation in Python.
2nd - make your implementation as fast as possible.
IF (capital letters because it's a big 'if') the code is still not fast enough and the cost/benefit of translating it to a compiled language is favourable THEN consider which compiled language to translate it in to. If you are in a field where Fortran is widely used, then learn Fortran by all means, but it is something of a niche language and it may benefit your career more to learn C++ or one of its relatives.
EDIT (too long to fit in the comment box)
Why mislead us in your question ? You state that the only language you are comfortable with is Python, now you say that you know Fortran. I guess you must be uncomfortable with it. And, from your comment, it seems that maybe the help you really need is in making your Python implementation faster; Sideshow Bob has offered some advice. Take that into consideration, then parallelise.
回答2:
A pointer towards a more efficient algorithm. If I am not mistaken you are comparing a point to every other point to see if any are closer than the linking length. For large numbers of points there are quicker ways of finding a near neighbours - spatial indexing and KD trees off the top of my head but doubtless there are other methods too that will work for you.
回答3:
If you have a modern graphics card you can paralelize to hundreds of processors (depending on your card) in Python code using PyOpenCL.
You can investigate to see if the algoritm FoF is implemented inside this voidfinder F90 code
You can define distance as the squared distance to avoid the use of sqrt() and use x*x instead of x**2...
来源：https://stackoverflow.com/questions/9333084/friends-of-friends-algorithm-written-in-python-need-to-be-in-fortran-90-95